Which of these two definitions is better or safer?

Question

char* varc = "ALICE";

char varl[] = "ALICE";

char* vars = (char*)calloc(6,sizeof(char));
strcpy(vars,"ALICE");

Which of the above definitions should I use more?

By the way, you [should not cast the result of `malloc`](https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) (or related functions like `calloc`). — Some programmer dude, Jan 11 '22 at 15:55

score 4 · Answer 1 · answered Jan 10 '22 at 16:40

char* varc = "ALICE";

This is a pointer to a CONSTANT (fixed) string, you cannot change the characters, so it should be:

const char* varc = "ALICE";

char varl[] = "ALICE";

This text can be changed after declaration, but it has a fixed size of 6-bytes.

char* vars = (char*)calloc(6,sizeof(char));
strcpy(vars,"ALICE");

With this one, you can choose the size of the buffer before putting any text in it, and you can over-allocate space if you might later want to put a longer string ("SAMANTHA") in it.

Overall, do not think in terms of "which one should I use more?".

Think in terms of "What are the limitations and abilities of each technique, and which one is correct for this specific use-case"

With the last one you have to remember to call `free(vars)` – Kevin Jan 10 '22 at 16:50 — Kevin, Jan 10 '22 at 16:50

Which of these two definitions is better or safer?

1 Answers1